Problem: Two lines defined by the equations $y = mx + 4$ and $y = 3x + b$, where $m$ and $b$ are constants, intersect at the point $(6, 10)$. What is the value of $b + m$?
Since the two lines intersect at $(6,10)$, each must pass through that point.  So, we can substitute this point for $x$ and $y$ in the two given equations, then solve for $m$ and $b$.  In the first equation we find: \begin{align*}
10&=m(6)+4\\
\Rightarrow\qquad 6&=6m\\
\Rightarrow\qquad 1&=m
\end{align*} Using the same method in the second equation gives: \begin{align*}
10&=3(6)+b\\
\Rightarrow\qquad 10&=18+b\\
\Rightarrow\qquad -8&=b
\end{align*} The value of $b+m$ is equal to $1+(-8)=\boxed{-7}$.